555 Blinking LED -- One cycle and stops

Discussion in 'The Projects Forum' started by ACRONYM, Jun 1, 2011.

  1. ACRONYM

    Thread Starter New Member

    Jun 1, 2011
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    Hi all,

    I built the circuit described here on my protoboard: http://wild-bohemian.com/electronics/images/blink-dia.gif

    R1: 56k ohms
    R2: 22k ohm resistor in series with a 1M variable resistor.
    C1: .1 uf

    I quadruple checked the wiring and it's all good, but the LED's don't strobe like they should. Instead, the LED's go through one cycle and stop. The LED going to ground flashes, then the other LED lights up and stays there.

    Thanks in advance
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Sounds like you don't have pins 2 and 6 connected together.

    What are you using for a power supply?

    Your LEDs would be flashing awfully quickly with the values you used.

    To get the flash durations of the two LEDs nearly equal, and slow the flash rate down to where you can see it (~12 pulses per second to about one pulse every 2 seconds), increase C1 to 1uF, change R1 to 2k to 5k Ohms (not less than 1.5k Ohms though), and leave the 56k and 1 MEG pot in series.

    Note that correct wiring for the pot to use as a rheostat is to connect the wiper (usually the middle terminal) and only one end to the circuit. If you connect the two "ends" without the wiper to the circuit, you won't be able to change the speed.
     
  3. k7elp60

    Senior Member

    Nov 4, 2008
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    Connect R1 between pins 8 and 7. Connect R2 between pins 6 and 7. Then the LEDs will flash back and forth and you can adjust the flash rate.
     
  4. ACRONYM

    Thread Starter New Member

    Jun 1, 2011
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    Thank you both, I tried both solutions but neither worked. I got the circuit to work earlier today, but it stopped working for no discernible reason.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    Well, protoboards go bad; especially if you have been using components with large leads (like TO-220 packages), or jumper wires larger than AWG-24.

    Try taking the circuit apart and re-building it on another area of your protoboard. It's only a few parts, so it should not take long to do.
     
  6. ACRONYM

    Thread Starter New Member

    Jun 1, 2011
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    Sgt, I moved the whole part to a new part of the board, one that I rarely use. No dice.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Are you using a CMOS 555 timer? As in, does it have a "C" in the 555 prefix? TLC555, ICM7555, etc are example part numbers.

    CMOS timers are vulnerable to static electricity discharges; you can "zap" them in a heartbeat and never feel the zap. It takes about 3,000 volts of static for you to feel it, but only about 25v-30v to kill a CMOS part.

    Also, the CMOS 555 timers can sink up to 100mA current, but can only source about 10mA. If you've been using a power supply higher than about (4.7v plus the rated Vf of your LEDs) you would have been overloading a CMOS 555 timer's output. Example: a red LED with a Vf of 2v, and a 7v supply with a 470 Ohm current limiter would be (7v-2v)/470 = 5/470 = 10.64mA, or about 106.4% of the rated output current.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    How about a photo? The schematic is sound, it will work if constructed correctly.
     
  9. ACRONYM

    Thread Starter New Member

    Jun 1, 2011
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    Top is ground, bottom is +6v
     
  10. Kingsparks

    Member

    May 17, 2011
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    It looks in the picture like pin 4 is NC. If so that is reset and should be tied to VCC when not in use.
     
  11. Kingsparks

    Member

    May 17, 2011
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    What colors are the resistors going from Vcc to pin 7 and from pin 7 to the pot? I think the pot. In the picture they looks like red, red, black and green blue black. Is that so? If so you have 220ohms and 560 ohms respectively in and you said you wanted 22K and 56K??? I think, or are my old eyes fooling me?
     
  12. Wendy

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    Mar 24, 2008
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  13. Kingsparks

    Member

    May 17, 2011
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    Bill.

    That's what I thought at first but I saved the photo, enhanced it and blew it up. As posted it is shown on the left edge of the board. When brightness is increased enough you can see the #1 marking, a indented dot, is on the upper right corner of the chip making what would be pin 3 actually pin 7. I think everything else is correct although the wires obscure a lot.

    I usually build my circuits with pin 1 oriented to the left so I thought it was that way until I noticed everything was off. No reason I guess just the way I do it.
     
    Last edited: Jun 2, 2011
  14. Wendy

    Moderator

    Mar 24, 2008
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    OK, the chip is backwards to what I thought it was. Pin 4 is floating, this is never a legal condition. It needs to go to Vcc.

    [​IMG]

    My Cookbook

    It would be clearer if it were laid out flat. In my projects I show how to lay something like this out, it is for real, not just a drawing.
     
  15. Kingsparks

    Member

    May 17, 2011
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    Yes, I agree and the way you show it is the way I draw all my 555 circuits. Once in a while I move pin 5 down to the bottom next to pin 1 but most of the time just the way you show it with pins 4 & 8 tied together and to Vcc unless I'm using the reset function.
     
  16. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    I think what Bill is trying to point out is that there is no connection on your protoboard to pin 4 of the 555 chip. The wire is missing.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    Take a look at what's connected to pin 3 - a single 100 Ohm resistor connected to a junction of two LEDs. That's not going to work.

    The two LEDs should not have been connected together.
    The two resistors R3 & R4 cannot be replaced by a single resistor.

    The timer IC is most likely burned up due to overloading the output. If the Vcc used was 9v, the poor timer was probably trying to source around 65mA current to LED2.

    Since our OP is using a CMOS timer, they will need to select resistors that will keep the current sunk by the 555 timer less than 100mA or the rated current of the LED that has its' anode connected to the positive rail (R3 for LED1), and less than 10mA for LED2 (R4).
    So:
    R3 needs to be >= (Vcc-Vf(LED1)) / 20mA
    For example: If Vf(LED1) = 2v, and Vcc=9v, then:
    R3 >= (9v-2v)/20mA = 7/0.02 = 350 Ohms. 360 Ohms is the closest standard E24 value.

    R4 needs to be >= (Vcc-Vf(LED2)) / 10mA
    For example: if Vf(LED2) = 2.5v, and Vcc=12v, then:
    R4 >= (12v-2.5v) / 10mA = 9.5/0.01 = 950 Ohms. 910 Ohms is the closest standard E24 value. 1k Ohms would also be OK to use.

    There is a graphical resistor color band decoder available on this page:
    http://samengstrom.com/24614782/en/read/4_Band_Resistor_Color_Codes

    There is a decade table of standard resistance values here:
    http://www.logwell.com/tech/components/resistor_values.html
    Use the E24 (green) or E12 (yellow) columns. The E48 and higher are more expensive. the E6 values are a subset of the E12 values; they can certainly be used, but the values are pretty coarse.
     
  18. Wendy

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  19. SgtWookie

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    Jul 17, 2007
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    Bill,
    Your R3 is pretty small at only 10 Ohms. If pot R4 were adjusted near 0, with a 9v supply, you would have pretty high instantaneous current going through R3 (~600mA, or 2/3 Vcc / 10 Ohms) when the 555 output changes states. That could cause the failure of pot R4.

    Changing C1 to 10uF, R4 to 100k and R3 to 100 Ohms would cut the instantaneous current to 60mA, the timing would be unchanged, and the capacitor would be smaller in physical size (if rated for the same voltage). Cap leakage could become more of an issue, but likely pretty inconsequential with fresh caps.
     
  20. ACRONYM

    Thread Starter New Member

    Jun 1, 2011
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    Again, thanks for the help.
    Kings, you are right and I made a terrible error, the resistors are 22 and 56 ohms.
    I found some 150k resistors and I put the 150k one in series with the 1M pot - it works wonderfully.
    Thanks again for the spot - problem solved.
     
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